Basis for Kernel and image of map

By definition$$\ker(f)=\left\{\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\in V\,\middle|\,f\left(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\right)=\begin{bmatrix}0\\0\end{bmatrix}\right\}.$$So, you can compute it by solving the system$$\left\{\begin{array}{l}x_1-x_3=0\\x_2-3x_3=0\end{array}\right.$$and searching for solutions in $V$. You will get that$$\ker(f)=\left\{\begin{bmatrix}t\\3t\\t\end{bmatrix}\,\middle|\,t\in K\right\}.$$It follows now from the rank-nullity theorem that $\dim\operatorname{Im}(f)=1$. So, and since$$f\left(\begin{bmatrix}1\\1\\0\end{bmatrix}\right)=\begin{bmatrix}1\\1\end{bmatrix},$$you have$$\operatorname{Im}(f)=\left\{\begin{bmatrix}t\\t\end{bmatrix}\,\middle|\,t\in K\right\}.$$Finally, it is clear that$$\left\{\begin{bmatrix}1\\3\\1\end{bmatrix}\right\}\quad\text{and}\quad\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}$$are bases of $\ker(f)$ and of $\operatorname{Im}(f)$ respectively.